Question

A charge +Q is placed at each of the diagonally opposite corners of a square. A charge -q is placed at each of the other diagonally opposite corners as shown. If the net electrical force on +Q is zero, then $\frac{-q}{+Q}$ is equal to

• A. +1 $\frac{+1}{\sqrt{2}}$
• B. +2$\sqrt{2}$
• C. -2$\sqrt{2}$

Answer 1

$\frac{-1}{2\sqrt{2}}$

Answer 2

To solve this problem, we need to consider the forces acting on one of the +Q charges due to the other charges. Let's denote the side length of the square as a.

1. Forces due to -q charges: Each -q charge exerts an attractive force on the +Q charge. The magnitude of each of these forces is given by Coulomb's law:

   $F = \frac{kQq}{a^2}$

   where k is Coulomb's constant. These forces are directed along the sides of the square towards the -q charges.

2. Force due to the other +Q charge: The other +Q charge exerts a repulsive force on the +Q charge. The distance between these charges is $\sqrt{2}a$ (the diagonal of the square). The magnitude of this force is:

   $F' = \frac{kQ^2}{2a^2}$

   This force is directed along the diagonal of the square, away from the other +Q charge.

3. Net Force: For the net force on +Q to be zero, the vector sum of the forces due to the two -q charges must be equal and opposite to the force due to the other +Q charge. The resultant force due to the two -q charges acts along the diagonal and has a magnitude of $\sqrt{2}F = \frac{\sqrt{2}kQq}{a^2}$.

   Therefore, we must have:

   $\frac{\sqrt{2}kQq}{a^2} = \frac{kQ^2}{2a^2}$

4. Solving for -q/+Q: Simplifying the equation, we get:

   $\sqrt{2}q = \frac{Q}{2}$

   $q = \frac{Q}{2\sqrt{2}}$

   Since we are looking for the ratio $\frac{-q}{+Q}$, and considering that the charges -q are negative, we have:

   $\frac{-q}{+Q} = -\frac{1}{2\sqrt{2}}$